长度为n的数组a初始值全为0，目标是把数组a变为数组b(1＜=bi＜=n), 可以进行任意次操作：选择长度为k的数组c，(1＜=ci＜=n且两两不同)

对于1<=i<=k, 把 a[c[i]] 改为c[i % k + 1]。给定n，k和数组b，判断能否得到数组b。

思路：

cpp 复制代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5;
const int mod = 1e9 + 7;
// const int mod = 998244353;
const int N = 1e6;
// int a[505][5005];
// bool vis[505][505];
// char s[505][505];
int a[maxn], b[maxn];
bool vis[maxn];
string s;
int n, m;

struct Node{
    int val, id;
    bool operator<(const Node &u)const{
        return val < u.val;
    }
    // int x, y;
    // int l, r, j;
}c[maxn];

int ans[maxn];

vector<int> G[maxn], g[maxn];
int dfn[maxn], low[maxn], in_st[maxn];
int col[maxn], cnt_col, tot;
int sum[maxn], siz[maxn];
stack<int> st;

void tarjan(int u){
	dfn[u] = ++tot;
	low[u] = dfn[u];
	in_st[u] = 1;
	st.push(u);
	for(auto v : G[u]){
    if(!dfn[v]){//v没遍历过，先更新v，更新u
			tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else{
			if(in_st[v]) //后向边或者横向边，就更新
			low[u] = min(low[u], dfn[v]);
		}
	}
 	if(low[u] == dfn[u]){//找到该强连通分量的最先访问到的点
		col[u] = ++cnt_col;//把该分量的点都标记为同一颜色
		sum[cnt_col] += a[u];
        siz[cnt_col]++;
        while(!st.empty() && st.top() != u){
            int x = st.top();
            st.pop();
            in_st[x] = 0;
            col[x] = cnt_col;
            sum[cnt_col] += a[x];
            siz[cnt_col]++;
        }
		if(!st.empty()){
			st.pop();
			in_st[u] = 0;
		}
	}
}

void solve(){
    int res = 0;
    int q, k;
    cin >> n >> k;
    
    tot = 0;
    cnt_col = 0;
    for(int i = 1; i <= n; i++){
        G[i].clear();
        g[i].clear();
        siz[i] = 0;
        in_st[i] = 0;
        dfn[i] = low[i] = 0;
    }
    while(!st.empty()){
        st.pop();
    }

    for(int i = 1; i <= n; i++){
        cin >> a[i];
        G[i].pb(a[i]);
    }
    if(k == 1){
        for(int i = 1; i <= n; i++){
            if(a[i] != i){
                cout << "No\n";
                return;
            }
        }
        cout << "Yes\n";
        return;
    }
    for(int i = 1; i <= n; i++){
        if(a[i] == i){
            cout << "No\n";//不能自环
            return;
        }
        if(!dfn[i]){
            tarjan(i);
        }
    }
    for(int i = 1; i <= cnt_col; i++){
        if(siz[i] > 1 && siz[i] != k){
            cout << "No\n";
            return;
        }
    }
    
    cout << "Yes\n";
    
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}

法二：

cpp 复制代码

#include <bits/stdc++.h>

using i64 = long long;

void solve() {
    int n, k;
    std::cin >> n >> k;
    
    std::vector<int> b(n);
    for (int i = 0; i < n; i++) {
        std::cin >> b[i];
        b[i]--;
    }
    
    if (k == 1) {
        for (int i = 0; i < n; i++) {
            if (b[i] != i) {
                std::cout << "NO\n";
                return;
            }
        }
        std::cout << "YES\n";
        return;
    }
    
    std::vector<int> vis(n, -1);
    for (int i = 0; i < n; i++) {
        int j = i;
        while (vis[j] == -1) {
            vis[j] = i;
            j = b[j];
        }
        if (vis[j] == i) {
            int len = 0;
            int x = j;
            do {
                len++;
                x = b[x];
            } while (x != j);
            if (len != k) {
                std::cout << "NO\n";
                return;
            }
        }
    }
    std::cout << "YES\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t;
    std::cin >> t;
    
    while (t--) {
        solve();
    }
    
    return 0;
}